Physics, asked by OoAryanKingoO79, 11 hours ago

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Answered by OoAryanKingoO78
11

Answer:

\orange{\boxed{\boxed{\begin{array}{cc}\bf \: \to \:given :   \\  \\  \hookrightarrow \:  \sf \:mass \: of \: the \: truck , \:  \: m = 4 \times  {10}^{7} \\  \\\hookrightarrow \:  \sf \:force ,  \:  \: F = 6 \times  {10}^{5} \:  N\\  \\ \hookrightarrow \:  \sf \:initial  \: velocity, \:  \: u = 0 \:  \:  \: [ \bf \: at \: rest]  \\  \\ \hookrightarrow \:  \sf \:time, \: t = 10 \: min \\  = 10 \times 60 \: s \\  = 600 \: s\\  \\  \\  \blue{ \underline{  \pink{\bf \: we \: have \: to \: find}}} \\  \\  \leadsto \:  \sf \: acceleration \: of \: the \: truck = a \\  \\  \small{ \leadsto \:  \sf \: final \: velocity \: of \: the \: truck \: after \: 10 \: min = v}\end{array}}}}

{\boxed{\boxed{\begin{array}{cc} \red{ \underline{\bf \: solution}} \\  \\  \text{we \: know \: that} , \\  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \sf \: F = ma \\ \\  \sf \:  \implies \:a =  \frac{F}{m}   \\  \\  =  \frac{6 \times  {10}^{5} }{4 \times  {10}^{ 7} }  \\  \\  =   \frac{3}{2} \times  {10}^{5- 7}   \\  \\  = 1.5 \times  {10}^{ - 2}  \\  \\  = 0.015 \: \sf \: m {s}^{ - 2}  \\  \\ \orange{ \boxed{\therefore \:  \sf \: acceleration, \: a = 0.015 \: m {s}^{ - 2} }}\end{array}}}}

Again,

{\boxed{\boxed{\begin{array}{cc}\bf \: we \: know \: that:  \\  \\  \sf \: v = u + at \\  \\  = 0 + 0.015 \times 600 \\  \\  =  \frac{1.5}{100} \times 600 \\  \\  = 1.5 \times 6 \\  \\  = 9 \: ms {}^{ - 1}   \\  \\  \orange{ \boxed{ \therefore \sf \: velocity, \:  \: v = 9 \: m {s}^{ - 1}}} \end{array}}}}

Answered by Dalfon
48

Question:

A truck of mass 4 × 10⁷ kg is pulled by a force of 6 × 10⁵ N. Calculate the acceleration. Find it's velocity after 10 minutes if initially the truck was at rest.

Answer:

a = 0.015 m/s², v = 9 m/s

Explanation:

As per question given that mass (m) = 4 × 10⁷ kg, force (f) = 6 × 10⁵ N, initial velocity (u) = 0 m/s, time (t) = 10 min = 10×60 = 600 sec

We need to find out the acceleration (a) and velocity (v).

Force = mass × acceleration

6 × 10⁵ = 4 × 10⁷ × a

(6 × 10⁵)/(4 × 10⁷) = a

0.015 = a

Hence, the acceleration is 0.015 m/s².

Now,

v = u + at

Substitute the values,

v = 0 + (0.015)(600)

v = 0 + 9

v = 9

Hence, the final velocity is 9 m/s.

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